The Azimuth Project
Experiments in Markov chains (Rev #23, changes)

Idea

This is page on Markov chains and Markov processes, focusing on their appearance in network theory.

The ethyl cation

Suppose you start with a molecule of ethane, which has 2 carbons and 6 hydrogens arranged like this:

Then suppose you remove one hydrogen. The result is a positively charged ion, or ‘cation’. This particular one is called an ‘ethyl cation’. People used to think it looked like this:

They also thought a hydrogen could hop from the carbon with 3 hydrogens attached to it to the carbon with 2. So, they drew a graph with a vertex for each way the hydrogens could be arranged, and an edge for each hop. It looked like this:

The red vertices come from arrangements where the first carbon has 2 hydrogens attached to it, and the blue vertices come from those where the second carbon has 2 hydrogens attached to it. So, each edge goes between a red vertex and a blue vertex.

This graph has 20 vertices, which are arrangements or ‘states’ of the ethyl cation. It has 30 edges, which are hops or ‘transitions’. Let’s see why those numbers are right.

First I need to explain the rules of the game. The rules say that the 2 carbon atoms are distinguishable: there’s a ‘first’ one and a ‘second’ one. The 5 hydrogen atoms are also distinguishable. But, all we care about is which carbon atom each hydrogen is bonded to: we don’t care about further details of its location. And we require that 2 of the hydrogens are bonded to one carbon, and 3 to the other.

So, there are 2 choices of which carbon has two hydrogens attached to it. Then there are (52)=10\binom{5}{2} = 10 choices of which two hydrogens are attached to it. This gives a total of 2×10=202 \times 10 = 20 states. These are the vertices of our graph.

The edges of the graph are transitions between states. The idea is that any hydrogen in the group of 3 can hop over to the group of 2. There are 3 choices for which hydrogen atom makes the jump. So, starting from any vertex in the graph there are 3 edges. This means there are 3×20/2=303 \times 20 / 2 = 30 edges. That’s 3 edges touching each of 20 vertices, but then to avoid double-counting we have to divide by 2, since each edge actually touches 2 vertices.

It’s very symmetrical. Clearly any permutation of the 5 hydrogens acts as a symmetry of the graph, and so does any permutation of the 2 carbons. This gives a symmetry group S5×S2S_5 \times S_2, which has 5!×2!=2405! \times 2! = 240 elements. And in fact this turns out to be the full symmetry group of the Desargues graph.

The Desargues graph, its symmetry group, and further applications to chemistry are discussed here:

The ethyl cation, revisited

If at any moment the state of the ethyl cation corresponds to some vertex of the Desargues graph, and it hops randomly along edges as time goes by, we can describe how this works using a random walk on the Desargues graph. This is a nice example of a Markov process. So, I’d like to say a bit more about this.

But first, an admission. After the paper by Balaban, Fǎrcaşiu and Bǎnicǎ came out, people gradually realized that the ethyl cation doesn’t really look like the drawing I showed you! In fact it’s a ‘nonclassical’ ion. In other words, its actual structure is not what you’d expect from the traditional ball-and-stick model where you take an ethane molecule, rip off a hydrogen and see what’s left. The ethyl cation really looks like this:

So, if we stubbornly insist on applying the Desargues graph to realistic chemistry, we need to find some other molecule to apply it to.

Trigonal bipyramidal molecules

Luckily, there are lots of options! They’re called trigonal bipyramidal molecules. They look like this:

The 5 balls on the outside are called ‘ligands’: they could be atoms or bunches of atoms. In chemistry, ‘ligand’ just means something that’s stuck onto a central thing. For example, in phosphorus pentachloride the ligands are chlorine atoms, all attached to a central phosphorus atom:

It’s a colorless solid, but as you might expect, it’s pretty nasty stuff: it’s not flammable, but it reacts with water or heat to produce toxic chemicals like hydrogen chloride.

Another example is iron pentacarbonyl, where 5 carbon-oxygen ligands are attached to a central iron atom:

You can make this stuff by letting powdered iron react with carbon monoxide. It’s a straw-colored liquid with a pungent smell!

Whenever you’ve got a molecule of this shape, the ligands come in two kinds. There are the 2 ‘axial’ ones, and the 3 ‘equatorial’ ones:

And the molecule has 2020 states… but only if count the states a certain way. We have to treat all 5 ligands as distinguishable, but think of two arrangements of them as the same if we can rotate one to get the other. The trigonal bipyramid has a rotational symmetry group with 6 elements. So, there are 5!/6=205! / 6 = 20 states.

The transitions between states are devilishly tricky. They’re called pseudorotations, and they look like this:

If you look carefully, you’ll see what’s going on. First the 2 axial ligands move towards each other to become equatorial. Then 2 of the 3 equatorial ones swing out to become axial. This fancy dance is called the Berry pseudorotation mechanism.

To get from one state to another this way, we have to pick 2 of the 3 equatorial ligands to swing out and become axial. There are 3 choices here. So, if we draw a graph with states as vertices and transitions as edges, it will have 20 vertices and 20×3/2=3020 \times 3 / 2 = 30 edges. That sounds suspiciously like the Desargues graph!

Puzzle 1. Show that the graph with states of a trigonal bipyramidal molecule as vertices and pseudorotations as edges is indeed the Desargues graph.

The 5-dimensional hypercube

Now I want to go back to the Desargues graph and sketch how to use some fancy math to study the Markov process it describes. For this it’s probably easiest to go back to our naive model of an ethyl cation:

Even though ethyl cations don’t really look like this, and we should be talking about some trigonal bipyramidal molecule instead, it won’t affect the math to come. Mathematically, the two problems are isomorphic! So let’s stick with this nice simple picture.

We can be a bit more abstract, though. A state of the ethyl cation is like having 5 balls, with 3 in one pile and 2 in the other. And we can focus on the first pile and forget the second, because whatever isn’t in the first pile must be in the second.

Of course a mathematician calls a pile of things a ‘set’, and calls those things ‘elements’. So let’s say we’ve got a set with 5 elements. Draw a red dot for each 2-element subset, and a blue dot for each 3-element subset. Draw an edge between a red dot and a blue dot whenever the 2-element subset is contained in the 3-element subset. We get the Desargues graph:

Now, I keep saying that we get this, but it would be quite a lot of work to check that it’s true. It’s actually less work to draw a big graph with vertices for all the subsets of our 5-element set! If we draw an edge whenever an nn-element subset is contained in an (n+1)(n+1)-element subset, the Desargues graph will be sitting inside this big graph.

Here’s what it looks like:

This graph has 252^5 vertices. It’s actually a picture of a 5-dimensional cube! The vertices are arranged in columns. There’s

• one 0-element subset,

• five 1-element subsets,

• ten 2-element subsets,

• ten 3-element subsets,

• five 4-element subsets,

• one 5-element subset.

So, the numbers of vertices in each column go like this:

15101051 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1

which is a row in Pascal’s triangle. We get the Desargues graph if we keep only the vertices corresponding to 2- and 3-element subsets, like this:

So, this is yet another way to draw the Desargues graph!

The graph Laplacian

Now, suppose that the process whereby our molecule hops from one state to another is ‘decoherent’ enough to be treated classically, rather than using quantum mechanics. Then we can describe this process as a random walk on the Desargues graph!

Note, all the transitions are reversible here: each edge in the Desargues graph really gives two edges in a directed graph, one pointing forward and one pointing backward. Thanks to the enormous amount of symmetry, the rates of all these transitions must be equal.

Let’s write DD for the set of vertices of the Desargues graph. A probability distribution of states of our molecule is a function

ψ:D→[0,1] \psi : D \to [0,1]

with

∑i∈Dψ=1 \sum_{i \in D} \psi = 1

We can think of these probability distributions as living in this vector space:

L1(D)={ψ:D→ℂ} L^1(D) = \{ \psi: D \to \mathbb{C} \}

I’m calling this space L1L^1 because of the general abstract nonsense explained in Part 5: probability distributions on any measure space XX live in a vector space called L1(X)L^1(X), consisting of the integrable functions on XX. Today the notation is overkill, since every function on DD is integrable. It’s just a finite set, after all, and these ‘integrals’ are just finite sums! But humor me, please.

The point is that we’ve got a general setup that applies here. There’s a Hamiltonian:

H:L1(D)→L1(D)H : L^1(D) \to L^1(D)

describing the rate at which the molecule randomly hops from one state to another… and the probability distribution Ψ∈L1(D)\Psi \in L^1(D) evolves in time according to the master equation

ddtΨ(t)=HΨ(t) \frac{d}{d t} \Psi(t) = H \Psi(t)

But what’s the Hamiltonian HH? It’s very simple, because it’s equally likely for the state to hop from any vertex x∈Dx \in D to any other vertex y∈Dy \in D that’s connected to xx by an edge. Lets write (yx)(y x) to mean that yy is connected to xx by an edge. Then

We’re subtracting 3Ψ(x)3 \Psi(x) because there are 3 edges coming out of each vertex xx, so this is the rate at which the probability of staying at xx decreases.

This Hamiltonian an example of a graph Laplacian. We could do the same thing for any graph, but it would get a tiny bit more complicated when different vertices have different numbers of edges coming out of them. Suppose GG is the set of vertices of a graph. Let n(x)n(x) be the number of vertices coming out of the vertex x∈Gx \in G. As before, write (yx)(y x) when there’s an edge connecting xx and yy. Then the graph Laplacian is this operator on L1(G)L^1(G):

There is a huge amount to say about graph Laplacians, but let’s just say that exp(tH)\exp(t H) describes time evolution for a random walk on the graph, where hopping from one vertex to any neighboring vertex has unit probability per unit time. We can make the hopping faster or slower by multiplying HH by a constant. (Here is a good time to admit that most people use a graph Laplacian that’s the negative of ours, and write time evolution as exp(−tH)\exp(-t H).

2) the off-diagonal entries of HH are nonnegative when we regard it as a G×GG \times G matrix in the obvious way.

This implies that for any t≥0t \ge 0, the operator exp(tH)\exp(t H) is stochastic—just what we want.

Exterior algebra

Now I’d like to give a cute description of the graph Laplacian that only works for the Desargues graph. Since this graph has 20 vertices, 10 red and 10 blue, the space L1(D)L^1(D) is 20-dimensional, the direct sum of two 10-dimensional pieces. But if you’re a mathematician, when you see these numbers

The symmetries of the Desargues graph are nicely evident now. S5S_5 acts to permute the basis vectors of ℝ5\mathbb{R}^5. S2S_2 switches the two summands; up to some signs it’s just the Hodge star operator!

I can’t resist adding that if we were studying our molecule quantum-mechanically instead of stochastically, we’d use a complex wavefunction instead of a real probability distribution, and we’d think of that as living in L2(D)L^2(D) instead of L1(D)L^1(D). But since DD is a finite set, L1(D)L^1(D) and L2(D)L^2(D) are the same. The big difference is that for the quantum problem must use a different Hamiltonian, since time evolution should be unitary, not stochastic.

And while we’re talking about quantum mechanics, I can’t resist adding that Λ(C5)\Lambda(\C^5) is fundamental to the description of fermions in the SU(5)\mathrm{SU}(5) grand unified theory! Each of the 32 basis vectors corresponds to some sort of quark, lepton, antiquark or antilepton. For more details, try:

I never expected this math to show up in chemistry! But when I saw it here, I got me an idea.

Recall how the exterior algebra Λ(ℂ5)\Lambda(\mathbb{C}^5) actually works. (If you don’t know, you can’t ‘recall’ tit, but I’m about to tell you, so you can pretend you knew.) The vector space ℂ5\mathbb{C}^5 has a standard basis, say e1,e2,e3,e4,e5e_1, e_2, e_3, e_4, e_5. We get a basis for Λ(ℂ5)\Lambda(\mathbb{C}^5) by taking ‘wedge products’ of these guys, for example

e1∧e3∧e2∧e4 e_1 \wedge e_3 \wedge e_2 \wedge e_4

and these wedge products satisfy some rules, most notably

ei∧ej=−ej∧ei e_i \wedge e_j = - e_j \wedge e_i

and therefore

ei∧ei=0 e_i \wedge e_i = 0

So, when we form our basis for Λ(ℂ5)\Lambda(\mathbb{C}^5), we only need to take wedge products of eie_i‘s in increasing order, since for example

and we don’t want the same eie_i to show up twice in one of these products, since then we’ll just get zero.

So, each eie_i can either be in the product, or not, so Λ(ℂ5)\Lambda(\mathbb{C}^5) has 252^5 different basis vectors, which is why it’s related to a 5-dimensional hypercube. And returning to our ethyl cation for a moment, we can say that each of these basis vectors corresponds to a state in which certain hydrogen atoms are attached to the first carbon atom. For example,

e1∧e2∧e5 e_1 \wedge e_2 \wedge e_5

corresponds to the state where the 1st, 2nd and 5th hydrogen are attached to the first carbon atom. The rest are attached to the second carbon atom.

By definition, Λp(ℂ5)\Lambda^p(\mathbb{C}^5) is the subspace whose basis consists of wedge products of pp different eie_i‘s. So, this subspace has dimension (5p)\binom{5}{p}. That’s why the dimensions of these subspaces give this row of Pascal’s triangle:

The aia_i terms take hydrogens off our carbon, and the ai†a_i^\dagger terms put them back on! Unfortunately this is wrong for several reasons.

For starters, in our ethyl cation problem, we want our carbon atom to have 2 or 3 hydrogens on it, not more or fewer. If it only has 2, we don’t want an annihilation operator to hit it and take away more! And if it has 3, we don’t want a creation operator to stick on more.

So, we need to think about how the space we’re really interested in sits inside the exterior algebra. There’s an inclusion

Sandwiching our original guess with ι\iota and ι†\iota^\dagger this way keeps us in the space of states with 2 or 3 hydrogens. Now the annihilation operators will give zero if we’re in a state with only 2 hydrogens, and the creation operators will give zero if we’re in a state with 3.

A commutative algebra

But there’s still a problem with using an exterior algebra, which I have concealed up to now. By definition, the wedge product is ‘anticommutative’:

These minus signs are no good in probability theory! A probability distribution has to be a nonnegative linear combination of our basis vectors. I hoped for a while that we could weasel around this somehow, but I don’t think we can.

So in fact the exterior algebra is the wrong thing! Instead we want a different algebra, namely

where ApA_p has a basis of consisting of products of pp different eie_i‘s, arranged in increasing order. The dimension of ApA_p matches that of Λp(ℂ5)\Lambda^p(\mathbb{C}^5): they’re isomorphic as vector spaces, but not canonically so, thanks to those darn minus signs.

In short, AA is isomorphic to Λ(ℂ5)\Lambda(\mathbb{C}^5) as a graded vector space, but not as an algebra. Nonetheless, we still have annihilation and creation operators defined almost as before:

still makes sense, but now the nasty minus signs are gone! All the matrix elements of HH are nonnegative now, with respect to the basis we’re using.

Unfortunately, there’s still one more problem: this operator HH is self-adjoint, which is great for quantum mechanics, but it’s not infinitesimal stochastic, so exp(tH)\exp(t H) won’t be stochastic.

Luckily this problem is easily fixed… and I think it’s the last problem. I think we just need to subtract 3 times the identity operator, much as we did when defining the Laplacian of the Desargues graph:

I claim that this operator is infinitesimal stochastic. Indeed I claim it’s just a disguised version of the graph Laplacian for the Desargues graph:

The vertices here correspond to the basis states ei∧eje_i \wedge e_j and ei∧ej∧eke_i \wedge e_j \wedge e_k (i<j<ki \lt j \lt k). The edges correspond to ways an annihilation or creation operator can map one basis state to another. And we need a −3I-3 I term in our Hamiltonian because there are 3 edges coming out of each vertex.

Puzzle 3. Show that this is true: if we use the obvious isomorphism to identify L1(D)L^1(D) with the space A2⊕A3A^2 \oplus A^3, the graph Laplacian of the Desargues graph equals

Conclusion

There’s much more to say, but I’ll be amazed if you’ve read this far, so let’s stop here. Anyway, if you ever actually see any iron pentacarbonyl or phosphorus pentachloride, you can now rejoice in the fact that it consists of many quadrillions of trigonal bipyramidal molecules, all doing pseudorotations, tracing out random walks on the Desargues graph.

One final thing. You may be curious as to why it’s called the ‘Desargues graph’. This name comes from Desargues’ theorem, a famous result in projective geometry. Suppose you have two triangles ABC and abc, like this:

Suppose the lines Aa, Bb, and Cc all meet at a single point, the ‘center of perspectivity’. Then the point of intersection of ab and AB, the point of intersection of ac and AC, and the point of intersection of bc and BC all lie on a single line, the ‘axis of perspectivity’! The converse is true too.

The Desargues configuration consists of all the actors in this drama:

• 10 points: A, B, C, a, b, c, the center of perspectivity, and the three points on the axis of perspectivity

Given any configuration of points and lines, we can form its Levi graph by drawing a vertex for each point or line, and drawing an edge whenever a point lies on a line. And now for the punchline: Levi graph of the Desargues configuration is the ‘Desargues-Levi graph’!—or Desargues graph, for short.

But I don’t know how this is relevant to anything I’ve discussed. For now it’s just a tantalizing curiosity.